时效性与时效性的时差? [英] Time difference in years with lubridate?

问题描述

我想用lubridate来计算给出他们的出生日期和今天的日期的年龄.现在我有这个:

I would like to use lubridate to calculate age in years given their date of birth and today's date. Right now I have this:

library(lubridate)
today<-mdy(08312015)
dob<-mdy(09071982)
today-dob

这给了我他们几天的年龄.

which gives me their age in days.

推荐答案

这是我要采用的lubridate方法:

interval(dob, today) / years(1)

得出32年的答案.

请注意，该函数将抱怨它无法表达该年剩余时间.这是因为年份不是固定的概念，即leap年为366，非-年为365.您可以获得有关周数和天数的更详细的答案:

Note that the function will complain that it cannot express the remainder of the fraction of the year. This is because year is not a fixed concept, i.e. 366 in leap years and 365 in non-leap years. You can get an answer with more detail in regard to the number of weeks and days:

interval_period = interval(dob, today)
full_year = interval_period %/% years(1)
remaining_weeks = interval_period %% years(1) %/% weeks(1)
remaining_days = interval_period %% years(1) %% weeks(1) %/% days(1)
sprintf('Your age is %d years, %d weeks and %d days', full_year, remaining_weeks, remaining_days)
# [1] "Your age is 32 years, 51 weeks and 1 days"

请注意，我使用%/%进行除法，并使用%%作为模数，以减去整年/周后的剩余周数/天.

Note that I use %/% for division and %% as modulo to get the remaining weeks/days after subtracting the full years/weeks.

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